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JEE Main 20266 AprilMorning Shift
Exam
JEE Main
Year
2026
Date
6 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main 2026 · 6 April · Morning Shift · Mathematics · Inverse Trigonometric Functions

Let 0 < α < 1 , β = (1)/(3α) and ^ -1 (1-α) + ^ -1 (1-β) = (π)/(4)...

Let 0<α<10 < \alpha < 1, β=13α\beta = \dfrac{1}{3\alpha} and tan⁡−1(1−α)+tan⁡−1(1−β)=π4\tan^{-1}(1-\alpha) + \tan^{-1}(1-\beta) = \dfrac{\pi}{4}. Then 6(α+β)6(\alpha + \beta) is equal to:

Options

  1. A66
  2. B77
  3. C88
  4. D99

Correct answer

B. 77

Step-by-step solution

Given tan⁡−1(1−α)+tan⁡−1(1−β)=π4\tan^{-1}(1-\alpha) + \tan^{-1}(1-\beta) = \dfrac{\pi}{4}

Applying the formula tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy)\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\dfrac{x+y}{1-xy}\right), we get:

(1−α)+(1−β)1−(1−α)(1−β)=tan⁡(π4)=1\dfrac{(1-\alpha) + (1-\beta)}{1 - (1-\alpha)(1-\beta)} = \tan\left(\dfrac{\pi}{4}\right) = 1

2−α−β=1−(1−α−β+αβ)2 - \alpha - \beta = 1 - (1 - \alpha - \beta + \alpha\beta)

2−α−β=α+β−αβ2 - \alpha - \beta = \alpha + \beta - \alpha\beta

2(α+β)=2+αβ2(\alpha + \beta) = 2 + \alpha\beta

Given β=13α\beta = \dfrac{1}{3\alpha}, we have αβ=13\alpha\beta = \dfrac{1}{3}. Substituting this into the equation:

2(α+β)=2+132(\alpha + \beta) = 2 + \dfrac{1}{3}

2(α+β)=732(\alpha + \beta) = \dfrac{7}{3}

α+β=76\alpha + \beta = \dfrac{7}{6}

Therefore, 6(α+β)=6×76=76(\alpha + \beta) = 6 \times \dfrac{7}{6} = 7

Answer: 77
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